I'm doing problem 2.10 in textbook Convex Optimization by Boyd and Vandenberghe:
Let $C \subseteq \mathbf{R}^{n}$ be the solution set of a quadratic inequality,
$$ C=\left\{x \in \mathbb{R}^{n} \mid x^{T} A x+b^{T} x+c \leq 0\right\} \\ $$
with $A \in \mathbb{S}^{n}, b \in \mathbb{R}^{n}$, and $c \in \mathbb{R}$.
(a) Show that $C$ is convex if $A \succeq 0$.
(b) Show that the intersection of $C$ and the hyperplane defined by $g^{T} x+h=0$ (where $g \neq 0)$ is convex if $A+\lambda g g^{T} \succeq 0$ for some $\lambda \in \mathbb{R}$.
Could you please verify if my solution for part (b) is correct?
Let $B = A+\lambda g g^{T}$. Then $A = B - \lambda g g^{T}$ and thus
$$ \begin{aligned} x^{T} A x+b^{T} x+c \leq 0 & \iff x^{T} (B - \lambda g g^{T}) x + b^{T} x+c \leq 0 \\ & \iff x^{T} B x - \lambda x^T g g^{T} x + b^{T} x+c \leq 0 \\ & \iff x^{T} B x - \lambda (g^T x)^T (g^{T} x) + b^{T} x+c \leq 0 \\ & \iff x^{T} B x - \lambda h^2 + b^{T} x+c \leq 0 \\ & \iff x^{T} B x + b^{T} x+(c - \lambda h^2) \leq 0 \end{aligned} $$
We have $B \succeq 0$ and is symmetric. Then it follows from (a) that $\{x \in \mathbb{R}^{n} \mid x^{T} B x + b^{T} x+(c - \lambda h^2) \leq 0 \}$ is convex. Moreover $\{x \in \mathbb{R}^{n} \mid g^{T} x+h=0\}$ is convex. Then the result follows as the intersection of convex sets is convex.
